Dirichlet form on Riemannian manifold is tight? $M$ is an $n$-dimensional Riemannian manifold. Consider the Dirichlet form $$\varepsilon \left( {u,v} \right) = \int_M {\left\langle {\nabla u,\nabla v} \right\rangle }, \quad u ,v \in {W^{1,2}}\left( M \right).$$  The capacity $c$ of an open set $A \subset M$ is defined by $$c\left( A \right): = \inf \{ {\left\| u \right\|_{{W^{1,2}}}}:u \ge 1 \text{ a.e. in $A$}\}. $$ And it can be extended to arbitrary sets $B \subset M$ by $$c\left( B \right): = \inf \{ c\left( A \right):B \subset A,A \text{ open}\}. $$ We say the Dirichlet form $\varepsilon $ is tight if there exist compact sets ${K_n} \subset M$, such that $c\left( {M\setminus {K_n}} \right) \to 0$. Is the Dirichlet form defined above on the Riemannian manifold tight?
 A: As discussed in comments, $W^{1,2}$ is interpreted here as $W^{1,2}_0$, the completion of $C^\infty_c(M)$ in the $W^{1,2}$ norm $\|f\|_{W^{1,2}}^2 = \int_M (f^2 + |\nabla f|^2)\,dVol$.  Also, the usual definition of capacity involves $\|u\|_{W^{1,2}}^2$ (your version is then the square root of this); I'll use the usual convention, though it doesn't affect the answer.
In general the capacity $c$ need not be tight (my comment was mistaken).  Consider the one-dimensional manifold $M = (0,1)$.  By Sobolev embedding (which has an elementary proof in this case), each function in $W^{1,2}$ is absolutely continuous and vanishes at the boundary $\{0,1\}$.  In particular, if $K$ is compact then there is no $f \in W^{1,2}$ with $f\ge 1$ on $(0,1) \setminus K$, so $c((0,1) \setminus K) = \infty$.
Indeed, $c$ is tight if and only if $1 \in W^{1,2}(M)$.  
For the forward direction, recall that since $\varepsilon$ is Dirichlet, if $f \in W^{1,2}$ then $f \wedge 1, f \vee 0 \in W^{1,2}$ as well.  If $c$ is tight then in particular there is a compact set $K$ with $c(M \setminus K) < \infty$, i.e. there exists $f \in W^{1,2}(M)$ with $f \ge 1$ a.e. on $M \setminus K$.  We can also find a function $g \in W^{1,2}$ with $g \ge 1$ on $K$ (indeed, we could take $g \in C^\infty_c(M)$ by a standard cutoff function construction).  Then $1 = ((f \vee 0) + (g \vee 0)) \wedge 1 \in W^{1,2}$.
Conversely, if $1 \in W^{1,2}(M)$ there is a sequence $f_n \in C^\infty_c(M)$ with $f_n \to 1$ in $W^{1,2}$-norm.  If $K_n$ is the support of $f_n$ then $g_n := 1 - f_n$ is a $W^{1,2}$ function with $g_n = 1$ on $M \setminus K_n$.  So $c(M \setminus K_n) \le \|g\|_{W^{1,2}} = \|1 - f_n\|_{W^{1,2}} \to 0$.
